Values of the second and third virial coefficients of nitrogen trifluoride are as follows: B = −94.3 cm^3·mol^−1 and C = 6740 cm^6.mol^−2 at 290 K B = −87.1 cm^3.mol^−1 and C = 6430 cm^6.mol^−2 at 300 K

Question

Values of the second and third virial coefficients of nitrogen trifluoride are as follows:

[latex] B=-94.3 cm ^3 \cdot mol ^{-1} ext { and } C=6740 cm ^6 \cdot mol ^{-2} ext { at } 290 K [/latex]

[latex] B=-87.1 cm ^3 \cdot mol ^{-1} ext { and } C=6430 cm ^6 \cdot mol ^{-2} ext { at } 300 K [/latex]

[latex] B=-80.4 cm ^3 \cdot mol ^{-1} ext { and } C=6090 cm ^6 \cdot mol ^{-2} ext { at } 310 K [/latex]

Calculate [latex]H^R and S^R [/latex] for nitrogen trifluoride at 300 K and 100 bar:

(a) Using expressions based upon the two-term virial equation.

(b) Using expressions based upon the three-term virial equation.

Accepted Answer

(a) Here, we employ Eqs. (6.55) and (6.56) for [latex]H_R and S_R[/latex], respectively. These require the temperature derivative of the second virial coefficient. For this, we use the values at 290 K and 310 K:

[latex] \frac{H^R}{R T}=\frac{P}{R}\left(\frac{B}{T}-\frac{d B}{d T} ight) [/latex] (6.55)

[latex] \frac{S^R}{R}=-\frac{P}{R} \frac{d B}{d T} [/latex] (6.56)

[latex] \frac{d B}{d T} \approx \frac{-80.4+94.3}{20}=0.695 cm ^3 \cdot mol ^{-1} \cdot K ^{-1} [/latex]

Using this value in Eq. (6.55) gives:

[latex] \frac{H^R}{R T}=\frac{P}{R}\left(\frac{B}{T}-\frac{d B}{d T} ight)=\frac{100}{83.14}\left(\frac{-87.1}{300}-0.695 ight)=-1.19 [/latex]

from which [latex]H^R = −2960 J·mol^−1.[/latex]

Similarly,

[latex] \frac{S^R}{R}=-\frac{P}{R} \frac{d B}{d T}=-\frac{100}{83.14}(0.695)=-0.836 [/latex]

from which [latex] S^R=-6.95 J \cdot mol ^{-1} \cdot K ^{-1} [/latex]

(b) In this case we require the temperature derivatives of both B and C, along with the value of the density of [latex]NF_3 [/latex] at 300 K and 100 bar. Solving the three-term virial coefficient iteratively as described in Ex. (3.8) gives[latex] V = 178.0 cm^3·mol^−1, or ρ = 0.00562 mol·cm^3.[/latex] This corresponds to a compressibility factor of Z = 0.714. Applying the same approach as in part (a) for the temperature derivative of C:

[latex] \frac{d C}{d T} \approx \frac{6740-6090}{20}=32.5 cm ^6 \cdot mol ^{-2} \cdot K ^{-1} [/latex]

With these values, Eqs. (6.62) and (6.63) give:

[latex] \frac{H^R}{R T}=T\left[\left(\frac{B}{T}-\frac{d B}{d T} ight) ho+\left(\frac{C}{T}-\frac{1}{2} \frac{d C}{d T} ight) ho^2 ight] [/latex] (6.62)

[latex] \frac{S^R}{R}=\ln Z-T\left[\left(\frac{B}{T}+\frac{d B}{d T} ight) ho+\frac{1}{2}\left(\frac{C}{T}+\frac{d C}{d T} ight) ho^2 ight] [/latex] (6.63)

[latex] \frac{H^R}{R T}=300\left[\left(\frac{-87.1}{300}-0.695 ight) 0.00562+\left(\frac{6430}{300}-\frac{32.5}{2} ight) 0.00562^2 ight]=-1.61 [/latex]

[latex] \frac{S^R}{R}=\ln (0.714)-300\left[\left(\frac{-87.1}{300}+0.695 ight) 0.00562+\frac{1}{2}\left(\frac{6430}{300}+32.5 ight) 0.00562^2 ight]=-1.27 [/latex]

from which [latex] H^R=-4021 J \cdot mol ^{-1} ext { and } S^R=-10.60 J \cdot mol ^{-1} \cdot K ^{-1} [/latex]

For comparison, values obtained from data in the NIST Chemistry WebBook are [latex] H^R=-3540 J \cdot mol ^{-1} ext { and } S^R=-8.83 J \cdot mol ^{-1} \cdot K ^{-1} [/latex]. The expression based on the two-term virial equation underestimates these, while the expression based on the three-term virial equation overestimates them. Note that the conditions selected correspond to [latex]T_r = 1.28 and P_r = 2.24.[/latex] These are outside the range where we would expect the two-term virial expansion to work well, and it ppears that adding a third term produces an overcorrection.

Question 6.4: Values of the second and third virial coefficients of nitrog...

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